Ms Access SQL calculation from multiple rows - sql

Question

I have the following query:
SELECT
InspectionID, Distance, Continuous, structural_grade
FROM
Conditions
WHERE
Continuous LIKE ("S%") OR Continuous LIKE ("F%")
ORDER BY
InspectionId ASC, Distance ASC
Which outputs:
262 60.80 S01 3
262 73.10 F01 3
262 82.60 S02 2
262 140.30 F02 2
263 30.80 S01 2
263 46.60 F01 2
380 4.60 S01 3
380 8.50 F01 3
380 9.80 S02 4
380 28.20 F02 4
380 77.70 S03 4
380 97.70 F03 4
The ordering is correct. However I'm stuck at what to do next.
Each "S" has a matching "F" with the same numbers Ex S01, F01. I need to access the Distance field of each "F" and minus "S" distance from it.
For examples F01 has a distance of 73.10 and S01 60.80 so the distance between is 12.3. I need to do this for each records and them numbers after the "S" and "F" can go really high yet will always match. ex. S999 - F999

Answer 1

Use the Mid() function to extract the digits from your Continuous field: Mid("S01", 2) yields "01". Then use those digits when you join the F and S rows.
With your sample data in Access 2010, the following query gave me this result set:
SELECT
f.InspectionID,
f.Continuous,
s.Continuous,
f.Distance AS F_distance,
s.Distance AS S_distance,
f.Distance - s.Distance AS F_minus_S
FROM
Conditions AS f
INNER JOIN
(
SELECT
InspectionID,
Continuous,
Distance,
Mid([Continuous],2) AS digits_only
FROM Conditions
WHERE Continuous ALike 'S%'
) AS s
ON
f.InspectionID = s.InspectionID
AND Mid(f.[Continuous],2) = s.digits_only
WHERE f.Continuous ALike 'F%'
ORDER BY f.InspectionId ASC, f.Distance ASC;
I included several columns which I doubt you want in your final query. But I think they can be useful during development and testing.
Beware the Access query designer will complain it can't represent that join condition in Design View. If you set up the join in SQL View, you can then run the query without Access complaining.

Answer 2

SELECT a.continuous AS 'S Continuous',
a.distance AS 'S Distance',
b.continuous AS 'F Continuous',
b.distance AS 'F Distance',
abs(a.distance - b.distance) AS 'Difference'
FROM Conditions a
INNER JOIN Conditions b
ON a.InspectionID = b.InspectionID
AND substring(a.continuous,2,len(a.continuous)-1) = substring(b.continuous,2,len(b.continuous)-1)
WHERE a.Continuous LIKE 'S%'
AND b.Continuous LIKE 'F%'
ORDER BY a.InspectionId ASC, a.Distance ASC